The buckling load for an axially loaded column with both ends fixed is given by:
Try it first: pick an option, then read the explanation.
Correct answer: C. 4π²EI/l²
Asked in SSC JE 2013
Detailed Explanation
Substitute the fixed–fixed effective length into Euler's formula.
\[ P = \frac{\pi^2EI}{(l/2)^2} = \frac{4\pi^2EI}{l^2} \]
Why “4π²EI/l²” Is Correct
The buckling load is 4π²EI/l².
| End conditions | Effective length le (Euler) | Crippling load P = π²EI/le² | Relative to hinged–hinged |
|---|---|---|---|
| Both ends hinged (pinned) | L | π²EI/L² | 1 |
| One end fixed, other free | 2L | π²EI/4L² | 1/4 |
| One end fixed, other hinged | L/√2 ≈ 0.707L | 2π²EI/L² | 2 |
| Both ends fixed | L/2 | 4π²EI/L² | 4 |
Longest effective length → smallest buckling load (fixed–free); shortest effective length → largest load (fixed–fixed).
Why the Other Options Are Wrong
- A. π²EI/l²: Hinged–hinged.
- B. 2π²EI/l²: Fixed–hinged.
- D. π²EI/(4l²): Fixed–free.
Key Point
Fixed–fixed: P = 4π²EI/l².
Key Concepts for Students
- Largest of the four standard Euler loads.
- le = l/2.
- Code design uses 0.65l.
Practice more: this is question 27 in Columns and Struts MCQs – Part 2. Try the whole set and check every answer instantly.